For the second part, the answer will be no in general. If one has a boundary which is a torus, and you require that the covering map be the elliptic involution, then the map won't extend in general over the 3-manifold, since this would extend to an involution on the manifold, which in general doesn't exist.
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Ian AgolJan 30 '11 at 19:09

1 Answer
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Berstein and Edmonds prove in Cor. 6.3 that for an orientable 3-manifold $W$ with connected boundary, with a branched cover $\varphi: \partial W\to S^2$ of degree $n>3$, then there is a branched cover $\Phi: W\to D^3$ such that $\Phi_{|\partial W}=\varphi$. In another paper, Edmonds claims in Theorem 2.1 that Cor. 6.3 extends to maps $f: W\to D^3$ such that the boundary map is a branched cover of the same degree as $f$ (allowable). One can easily construct an allowable map $f:W\to D^3$ by mapping $\partial W$ to $S^2$ by a branched cover so that each component of $\partial W$ has positive degree $>2$ (with respect to the orientation induced by $W$), and extend to all of $W$ by coning off. Theorem 2.1 implies that this map is homotopic to a branched cover.

The hypothesis of degree $>2$ is necessary, since for example if one has a knot $K\subset S^3$ which is not (strongly) invertible, then $M=S^3-\mathcal{N}(K)$ is a manifold with torus boundary such that there is a degree 2 map $T^2=\partial M\to S^2$ which is the quotient of the elliptic involution, but which doesn't extend over $M$.